Question

The A-36 steel bolt is tightened within a hole so that the reactive torque on the shank AB can be expressed by the equation t = (kx2) N.m/m, where x is in meters. If a torque of T = 50 N.m is applied to the bolt head, determine the constant k and the amount of twist in the 50-mm length of the shank. Assume the shank has a constant radius of 4 mm. The shear modulus of elasticity for A-36 steel is 75 GPa.

Answer 1

Answer:$k=1.2\times 10^6 N/m^2,\theta =3.56^{\circ}$

Explanation:

Given

$t=kx^2 Nm/m$

And a torque of 50 N.m is applied thus

$T=\int_{0}^{0.05}kx^2dx$

$50=\frac{1}{3}\left [ 5\times 10^{-2}\right ]^3$

$k=1.2\times 10^6 N/m^2$

T at a distance x is given

$T=\int_{0}^{x}1.2\times 10^6x^2dx$

$T=0.4\times 10^6.x^3$

For Angle of twist

$\theta =\frac{TL}{GJ}$

$J=\frac{\pi d^4}{32}$

$J=\frac{\pi \times 8^4\times 10^{-12}}{32}=0.402176$

G=75 GPa

$\theta =\int_{0}^{0.05}\frac{\left ( 50-0.4\times 10^6x^3\right )}{GJ}$

$\theta =3.56^{\circ}$